Prove that $(\mathbb{Z},+) \text{ and } (\mathbb{Q},+)$ are not isomorphic.
Usually in order to show that two groups are not isomorphic I would show that either the groups have different orders, one is abelian and the other isn't, one is cyclic and the other isn't or one contains an element of order d and the other doesn't. I can see that this isn't going to work here though.
I feel like a proof by contradiction is the way to go with this, I'm not sure how to proceed with that in this case though.
Uri Goren has the right of it, but I thought it worth elaborating a bit if you're still getting the hang of this sort of argument.
As you've essentially noticed already, you can prove this by observing that $(\mathbb{Q},+)$ has a "group theoretic property" that $(\mathbb{Z},+)$ does not. In other words, identify a statement about $(\mathbb{Q},+)$ that can't possibly be preserved by an isomorphism $\varphi: \mathbb{Q} \to \mathbb{Z}$ of additive groups. Formalizing such a proof is often done either (a) by contradiction or (b) by showing no homomorphism $\varphi: \mathbb{Q} \to \mathbb{Z}$ exists that is surjective (or injective).
For instance, suppose $\varphi: (\mathbb{Q},+) \to (\mathbb{Z},+)$ is a homomorphism. Then $\varphi(1)$ is an integer, as is $\varphi(1/n)$ for all $n \geq 1$, and you can show that
$$ \varphi(1/n) = \varphi(1)/n. $$
Thus $\varphi(1)/n \in \mathbb{Z}$ for all $n \in \mathbb{N}$, so what is $\varphi(1)$? What can you conclude about $\varphi(1/n)$ and $\varphi(m/n)$ accordingly?